Leysa y=tan(pix)

$$y= \tan ( \pi x )$$

Leystu fyrir x

$$x=2n_{1}+\frac{\arcsin(\frac{y}{\sqrt{y^{2}+1}})+\pi }{\pi }\text{, }n_{1}\in \mathrm{Z}\text{, }\exists n_{3}\in \mathrm{Z}\text{ : }\left(n_{1}>\frac{2n_{3}-\frac{2\arcsin(\frac{y}{\sqrt{y^{2}+1}})}{\pi }-1}{4}\text{ and }n_{1}<\frac{2n_{3}-\frac{2\arcsin(\frac{y}{\sqrt{y^{2}+1}})}{\pi }+1}{4}\right)$$

$$x=2n_{2}+\frac{\arcsin(\frac{y}{\sqrt{y^{2}+1}})}{\pi }\text{, }n_{2}\in \mathrm{Z}\text{, }\exists n_{3}\in \mathrm{Z}\text{ : }\left(n_{3}>\frac{4n_{2}+\frac{2\arcsin(\frac{y}{\sqrt{y^{2}+1}})}{\pi }-3}{2}\text{ and }n_{3}<\frac{4n_{2}+\frac{2\arcsin(\frac{y}{\sqrt{y^{2}+1}})}{\pi }-1}{2}\right)$$

Leystu fyrir y

$$y=\tan(\pi x)$$

$$\nexists n_{1}\in \mathrm{Z}\text{ : }x=n_{1}+\frac{1}{2}$$